5 Best Ways to Find the Minimum Difference Between Elements From Two Lists in Python

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💡 Problem Formulation: Given two lists of numerical values, how can one find the minimum absolute difference between any two elements, where one element comes from the first list and the other from the second? For instance, if we have lists [1, 3, 15] and [23, 127, 235], the minimum difference is between 15 and 23, which is 8.

Method 1: Brute Force Comparison

This method involves comparing each element from the first list with every element in the second list to find the absolute minimum difference. It’s straightforward but not efficient for large lists as it has a time complexity of O(n*m), where n and m are the lengths of the lists.

Here’s an example:

def min_diff_brute_force(list1, list2):
    min_diff = float('inf')
    for i in list1:
        for j in list2:
            diff = abs(i - j)
            if diff < min_diff:
                min_diff = diff
    return min_diff

# Example usage
list1 = [1, 3, 15]
list2 = [23, 127, 235]
print(min_diff_brute_force(list1, list2))

Output: 8

This function takes each element of the first list and subtracts it from every element of the second list, updating the minimum difference found. The time taken can be substantial for large lists, making it impractical in those cases.

Method 2: Sort and Compare

By sorting both lists, we can then iterate through them in tandem to find the minimum difference more efficiently. This method has a time complexity of O(n*log(n) + m*log(m)), which is the time needed to sort the lists.

Here’s an example:

def min_diff_sort_and_compare(list1, list2):
    list1, list2 = sorted(list1), sorted(list2)
    i, j, min_diff = 0, 0, float('inf')
    
    while i < len(list1) and j < len(list2):
        diff = abs(list1[i] - list2[j])
        if diff < min_diff:
            min_diff = diff
        if list1[i] < list2[j]:
            i += 1
        else:
            j += 1

    return min_diff

# Example usage
list1 = [1, 3, 15]
list2 = [23, 127, 235]
print(min_diff_sort_and_compare(list1, list2))

Output: 8

This algorithm first sorts both lists, then scans through them, incrementally advancing in one of the lists to find the smallest difference efficiently due to the sorted order.

Method 3: Using Heaps

Here, we utilize a min-heap data structure to keep all elements from one list and iteratively find the closet elements to each entry of the other list. While better organized, the performance is similar to the Brute force method, with O(n*log(n) + m*log(n)) complexity due to heap operations.

Here’s an example:

import heapq

def min_diff_heap(list1, list2):
    heapq.heapify(list1)
    min_diff = float('inf')

    for element in list2:
        closest = heapq.nsmallest(1, list1, key=lambda x: abs(x-element))
        min_diff = min(min_diff, abs(element - closest[0]))

    return min_diff

# Example usage
list1 = [1, 3, 15]
list2 = [23, 127, 235]
print(min_diff_heap(list1, list2))

Output: 8

The code converts the first list into a heap and then, for each element in the second list, finds the closest element in the heap, updating the minimum difference as needed.

Method 4: Binary Search

Upon sorting one list, we can perform a binary search for each element of the unsorted list within the sorted list. This way, we find the closest elements with a better time complexity, particularly O(n*log(m)) after one initial list sort.

Here’s an example:

import bisect

def min_diff_binary_search(sorted_list, other_list):
    min_diff = float('inf')
    
    for element in other_list:
        pos = bisect.bisect_left(sorted_list, element)
        if 0 < pos:
            min_diff = min(min_diff, abs(element - sorted_list[pos-1]))
        if pos < len(sorted_list):
            min_diff = min(min_diff, abs(element - sorted_list[pos]))
    
    return min_diff

# Example usage
list1 = [1, 3, 15]
list2 = [23, 127, 235]
sorted_list1 = sorted(list1)
print(min_diff_binary_search(sorted_list1, list2))

Output: 8

The function uses Python’s bisect module to find the insertion point for each element in other_list into the sorted list, checking the nearest elements for the smallest difference.

Bonus One-Liner Method 5: Utilizing itertools and min

This approach uses Python’s itertools to generate all combinations of pairs between the lists and find the minimum difference using a one-liner with a generator expression inside the min() function. It is concise but has the same O(n*m) complexity as the brute force approach.

Here’s an example:

from itertools import product

list1 = [1, 3, 15]
list2 = [23, 127, 235]
min_diff = min(abs(a - b) for a, b in product(list1, list2))
print(min_diff)

Output: 8

The single line of code generates all possible pairs using product from the itertools module and finds the pair with the smallest absolute difference.

Summary/Discussion

  • Method 1: Brute Force Comparison. Easy to understand and implement. It’s inefficient for large lists.
  • Method 2: Sort and Compare. Efficient for moderately sized lists. Requires additional space to sort lists, but time complexity is reduced due to sorting.
  • Method 3: Using Heaps. Offers organized data structure benefits but still involves substantial computation, similar to the brute force approach.
  • Method 4: Binary Search. Highly efficient for large lists, especially when the difference in sizes between the two lists is significant.
  • Bonus One-Liner Method 5: Very concise, suitable for small lists or as a simple script solution. Not efficient for large datasets.